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Math Help - Abstract Algebra- Pre-Exam

  1. #1
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    Abstract Algebra- Pre-Exam

    Well , I'm trying to solve some exams, and I have some difficulties in the following questions:

    1. Let A = where p,q are disjoint primes.
    Prove that there are unique m,n such as n|m and A is isomorphic to:
    .

    I've no clue about this one.

    2. Let F be a field.
    Prove that  \frac{F[x_{1},x_{2}]}{(x_{1})} isomorphic F[x_{2}]

    I realy need your guidance in these ones.

    Thanks in advance
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  2. #2
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    Quote Originally Posted by WannaBe View Post
    Well , I'm trying to solve some exams, and I have some difficulties in the following questions:

    1. Let A = where p,q are disjoint primes.
    Prove that there are unique m,n such as n|m and A is isomorphic to:
    .

    I've no clue about this one.
    The answer is n=p, m=p^2q. Search google with "torsion coefficients" or "invariant factors" for finitely generated abelian groups. I'll give another example. For \mathbb{Z}_6 \times \mathbb{Z}_{12} \times Z_{20} \cong \mathbb{Z}_{2} \times \mathbb{Z}_{3} \times \mathbb{Z}_{3}\times \mathbb{Z}_{4} \times \mathbb{Z}_{4} \times \mathbb{Z}_{5} , you can decompose like,

    \begin{matrix}<br />
2&4&4\\<br />
&3&3\\<br />
&&5 \end{matrix}.

    Torsion coefficients (multiply numbers in each column) are 2, 12, 60 where 2|12 and 12|60.

    2. Let F be a field.
    Prove that  \frac{F[x_{1},x_{2}]}{(x_{1})} isomorphic F[x_{2}]

    I realy need your guidance in these ones.

    Thanks in advance
    Let f:F[x_1, x_2] \rightarrow F[x_2], where x_1 \mapsto 0, x_2 \mapsto x_2. You can check that f is a surjective group homomorphism. Now the kernel of f is (x_1), the result follows from the first isomorphism theorem.
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  3. #3
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    Thanks! Just to make sure:
    In  F[x_{1},x_{2}] , every element has the form:
     f(x)*x_{1}+g(x)*x_{2} , right?


    Thanks again!
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  4. #4
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    Quote Originally Posted by WannaBe View Post
    Thanks! Just to make sure:
    In  F[x_{1},x_{2}] , every element has the form:
     f(x)*x_{1}+g(x)*x_{2} , right?


    Thanks again!
     F[x_{1},x_{2}] is the ring of polynomials in two variables x_1, x_2 over F. x is not a variable in this ring.
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  5. #5
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    Yep, you're right...So what's the form of the poly. in this ring?

    Thanks!
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  6. #6
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    Quote Originally Posted by WannaBe View Post
    Yep, you're right...So what's the form of the poly. in this ring?

    Thanks!
    F[x_1, x_2]=F[x_1][x_2]=\{\sum_{i,j>=0}a_{i,j}x_1^ix_2^j |a_{ij} \in F\}, where a_{ij}=0 all but finite number of values of i and j.
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  7. #7
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    Subring without identity?

    oops, sorry, posted something in the wrong place.
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  8. #8
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    thanks a lot!
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